All categories

3 years ago

Solve the following equation algebraically: 3 x squared = 12

Mathematics

2 answers:

Inessa05 [86]3 years ago

7 0

X = 2 or -2

either way it’s still 12 , hope this helps !

BabaBlast [244]3 years ago

3 0

Answer: x = ± 2

Step-by-step explanation:

3x² = 12

x² = 12/3

x² = 4

x = √4

x = ± 2

Answer: x = ± 2

<h2><em>Spymore</em></h2>

You might be interested in

(0.a * 10) - (0.a ÷ 10) = 6.93

marin [14]

Answer: please clarify

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Express 0.0939 as a fraction

DaniilM [7]

939/10,000 the decimal place indicates the fraction. You would love the decimal over 4 spaces which is the ten thousands place so that is what you set 939 over!

3 0

2 years ago

Read 2 more answers

Please help with question A and show work!! thank you

Digiron [165]

The equation for the line of reflection is x = 2.
If you drew a vertical line crossing the x-axis on x = 2, the drawing will be the same on the left and right side, as if one side was drawn then exactly copied on the other side but flipped.
The figure is symmetrical, and x = 2 is the line of symmetry.
Hope this helps!

8 0

3 years ago

1) There are 43 girls and 25 boys in the Valley View Middle School chorus. What is the ratio of girls to boys? 43 to 25, or 43:2

Arturiano [62]

Answer:

Oh gosh. Im not positive but I think it would be 43:25. Please dont get mad at me if its wrong!

Step-by-step explanation:

I am so sorry if its wrong but im not very sure :-(

6 0

2 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago